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AIC with ALSs
A Locked Set is a group of cells (that can all see each other) of size N where the number of candidates in those cells is equal to the size of the group. That is N cells contain N candidates. A solved cell or a clue is a Locked Set where N=1, but such a cell is not useful. The smallest useful Locked Set is a Naked Pair (where N=2) followed by a Naked Triple (N=3).

An Almost Locked Set (ALS) is N cells containing N+1 candidates. In the context of Alternating Inference Chains in this solver, an ALS is of size N=2 and the number of different candidates in those cells is 3, although bigger ALS groups are possible. So an ALS of size 2 will be a two Conjugate Pairs plus one other candidate.

 This might be easier to explain with a diagram. The highlighted yellow cells D7 and D8 contain [2,7] and [2,7,9] respectively. If D8 didn't contain a 9 the two cells would be a Naked Pair. The extra 9 makes these two cells an ALS.While solving a puzzle I am hunting around for Inference Chains and perhaps I start with the bi-value cell J2 containing [2,9]. I pretend to turn ON the 2 which turns OFF the 9 in that cell. The only other 9 is in J8 which I turn ON. Looking up the column the 9 in D8 must be turned OFF. This creates an on-the-fly Naked Pair. Now, a Naked Pair eliminates candidates in the row or column (or box) it is aligned on so we can use this elimination property as part of our chain. This is the trick! By removing the 9 in D8 we fix 2 and 7 into those two cells so we can look along the row at other 2s and 7 and turn them OFF. This I do in cell D3. From there I can continue the inference chain. ALS in an AIC fragment
 A real life example now. This chain contains an ALS on the cells {G6,H6} (I used squiggly brackets to denote ALS as opposed to square brackets for Grouped Cells). 9s in row H are the entry point. We turn 9 ON in H2 which turns OFF the 9 in H6 - the extra candidate that makes the ALS an ALS. This gives us a Naked Pair of [5,7] that points up column 6 turning OFF the 7 in F6 and the chain continues. Ultimately we use Nice Loop Rule 2 to place 4 in A4AIC on 4 (Discontinuous Alternating Nice Loop, length 12):-4[A4]+4[D4]-7[D4]+7[D2]-7[H2]+9[H2]-9[H6]+7{H6|G6}-7[F6]+4[F6]-4[A6]+4[A4]- Contradiction: When 4 is removed from A4 the chain implies it must be 4 - other candidates 2/5 can be removed AIC with ALS: Load Example or : From the Start
 This second example uses a chain to kill off-chain candidates, which is Nice Loop Rule 1. The ALS is in {F1,F4} and consists of [1/3/8] and [1/3] respectively. We turn off the extra candidate, 8 in F1 to enable the Naked Pair to be formed. Off Chain eliminations: Load Example or : From the Start
Alternating Inference Chain
AIC Rule 1: -3[B5]+6[B5]-6[B8]+6[D8]-8[D8]+8[D1]-8[F1]+3{F1|F4}-3[F3]+3[B3]-3[B5]
- Off-chain 6 taken off B9 - weak link: B5 to B8
- Off-chain candidates 1 taken off cell D8, link is between 6 and 8 in D8
- Off-chain 8 taken off F2 - weak link: D1 to F1
- Off-chain 8 taken off F3 - weak link: D1 to F1
- Off-chain 8 taken off J1 - weak link: D1 to F1
- Off-chain 3 taken off B4 - weak link: B3 to B5

 Go back to AICs with Grouped Cells Continue to Digit Forcing Chains

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## Tuesday 4-Jun-2013

### ... by: Nono

Question like Str8tsFan.
the last solver version 1.95 does not eliminate the 8 in J1.
No problem for good sudoku players !

Andrew Stuart writes:

Quite correct. There was some very odd code in there that stopped J1 from being eliminated because the link had already eliminated in the box. Can't say why, must have had a brain fart that day. Diagram and solver updated.

## Wednesday 3-Oct-2012

### ... by: Str8tsFan

(1) Using the link to the solver leads to a sudoku with a tiny little difference: B9 has an additional candidate 6, which is missing at the example. At the solver this candidate will be eliminated with the very same example:
"- Off-chain candidate 6 taken off B9 - weak link: B5 to B8"

(2) What about the candidate 8 at J1? As far as I can understand the theory, the weak link "D1 to F1" should not only eliminate the 8s at F2 and F3 (weak link in same box) but also the 8 at J1 (weak link in same column). More interesting: the solver doesn't eliminate that 8 at J1 either. Why? Did I make any mistake, or is it a flaw of the solver? As far as I understood, a weak link can be part of two entities (box and row or box and column) and thus should be able to eliminate candidates at both entities, maybe the solver fails to check the second entity?

Andrew Stuart writes:

Thanks for the detailed hints. Both points correct. I've updated the diagram and solver.

## Tuesday 10-Jul-2012

### ... by: Mr Turner

The last paragraph seems flawed since there is an 8 at F8. The 8s at F2 and F3 could be colored off as an inference from D1 being on. This implies F8 is on. F8 is also inferred to be on from D8 being off.

Andrew Stuart writes:

Correct. I've removed that paragraph now. Doh.

## Wednesday 22-Apr-2009

### ... by: Mr Archibald

I call this the 6 pack.

the first three in a line on rows 1,2 is the some as the last six on row 3. this can work upsidedown and backtofront and sideways.

a b c d e f g h i
i h g f e d c b a
d e f a b c i h g